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Standard heat of formation, [tex]\Delta H_{i}[/tex], is defined as the enthalpy change for the formation of one mole of a substance from its constituent elements in their standard states. Thus, elements in their standard states have [tex]\Delta H_i = 0[/tex]. Heat of formation values are used to calculate the enthalpy change of any reaction.

Consider, for example, the reaction:
[tex]\[ 2 NO (g) + O_2(g) \rightleftharpoons 2 NO_2(g) \][/tex]

The heat of formation values is given by the following table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
Substance & \begin{tabular}{c}
$\Delta H_{f}^{\circ}$ \\
$(kJ/mol)$
\end{tabular} \\
\hline
$NO (g)$ & 90.2 \\
\hline
$O_2(g)$ & 0 \\
\hline
$NO_2(g)$ & 33.2 \\
\hline
\end{tabular}
\][/tex]

The standard heat of reaction for the overall reaction is:
[tex]\[
\begin{aligned}
\Delta H_{\text{rxn}} &= \Delta H_i \text{ (products)} - \Delta H_i \text{ (reactants)} \\
&= 2(33.2) - 2(90.2) \\
&= -114 kJ
\end{aligned}
\][/tex]

Part A

For which of the following reactions is [tex]\Delta H_{m}^\circ[/tex] equal to [tex]\Delta H_{i}^\circ[/tex] of the product(s)? You do not need to look up any values to answer this question. Check all that apply.

1. [tex]Na (s) + \frac{1}{2} F_2 (l) \rightarrow NaF (s)[/tex]
2. [tex]CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g)[/tex]
3. [tex]CO (g) + \frac{1}{2} O_2 (g) \rightarrow CO_2 (g)[/tex]
4. [tex]2 Na (s) + F_2 (g) \rightarrow 2 NaF (s)[/tex]
5. [tex]Na (s) + \frac{1}{2} F_2 (g) \rightarrow NaF (s)[/tex]
6. [tex]C (s, \text{graphite}) + O_2 (g) \rightarrow CO_2 (g)[/tex]

Submit

Hints:
- Look again at the states of the reactants.
- The heat of formation is defined by a reaction in which the reactants are elements in their standard states.
- Review Hint 1: How to approach the problem.


Sagot :

To determine for which of the given reactions the enthalpy change, [tex]\(\Delta H_m^\rho\)[/tex], is equal to the standard enthalpy of formation, [tex]\(\Delta H_i^\rho\)[/tex], of the product(s), we need to identify the reactions where the reactants are in their standard states and form the products directly.

The definition of the standard enthalpy of formation, [tex]\(\Delta H_i^\rho\)[/tex], is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. Elements in their standard states have an enthalpy of formation, [tex]\(\Delta H_i^\rho\)[/tex], of zero.

Let's analyze each reaction one by one:

1. Na (s) + [tex]\(\frac{1}{2}\)[/tex] F[tex]\(_2\)[/tex] (l) [tex]\(\rightarrow\)[/tex] NaF (s)
- Sodium (Na) is in its standard state (solid).
- Fluorine (F[tex]\(_2\)[/tex]) is in the liquid state, but its standard state is gaseous.
- Hence, the reactants are not all in their standard states.
- This reaction does not meet the criteria.

2. CaCO[tex]\(_3\)[/tex] (g) [tex]\(\rightarrow\)[/tex] CaO + CO[tex]\(_2\)[/tex] (g)
- This is a decomposition reaction, breaking down calcium carbonate into calcium oxide and carbon dioxide.
- It’s not forming a compound from elements in their standard states.
- This reaction does not meet the criteria.

3. CO (g) + [tex]\(\frac{1}{2}\)[/tex] O[tex]\(_2\)[/tex] (g) [tex]\(\rightarrow\)[/tex] CO[tex]\(_2\)[/tex] (g)
- Carbon monoxide (CO) and oxygen (O[tex]\(_2\)[/tex]) are both in their standard gaseous states.
- The reaction forms carbon dioxide (CO[tex]\(_2\)[/tex]) directly.
- This reaction meets the criteria.

4. 2 Na (s) + F[tex]\(_2\)[/tex] (g) [tex]\(\rightarrow\)[/tex] 2 NaF (s)
- Sodium (Na) is in its standard state (solid), and fluorine (F[tex]\(_2\)[/tex]) is in its standard state (gas).
- However, this reaction involves the formation of two moles of NaF, not one mole.
- This reaction does not meet the criteria strictly as per the standard formation definition.

5. Na (s) + [tex]\(\frac{1}{2}\)[/tex] F[tex]\(_2\)[/tex] (g) [tex]\(\rightarrow\)[/tex] NaF (s)
- Sodium (Na) is in its standard state (solid), and fluorine (F[tex]\(_2\)[/tex]) is in its standard state (gas).
- The reaction forms one mole of NaF.
- This reaction meets the criteria.

6. C (s, graphite) + O[tex]\(_2\)[/tex] (g) [tex]\(\rightarrow\)[/tex] CO[tex]\(_2\)[/tex] (g)
- Carbon in its graphite form is in its standard state (solid), and oxygen (O[tex]\(_2\)[/tex]) is in its standard state (gas).
- The reaction forms carbon dioxide (CO[tex]\(_2\)[/tex]) directly.
- This reaction meets the criteria.

Therefore, the reactions for which [tex]\(\Delta H_m^\rho\)[/tex] is equal to [tex]\(\Delta H_i^\rho\)[/tex] of the product(s) are:

- CO (g) + [tex]\(\frac{1}{2}\)[/tex] O[tex]\(_2\)[/tex] (g) [tex]\(\rightarrow\)[/tex] CO[tex]\(_2\)[/tex] (g)
- Na (s) + [tex]\(\frac{1}{2}\)[/tex] F[tex]\(_2\)[/tex] (g) [tex]\(\rightarrow\)[/tex] NaF (s)
- C (s, graphite) + O[tex]\(_2\)[/tex] (g) [tex]\(\rightarrow\)[/tex] CO[tex]\(_2\)[/tex] (g)